插入一對一的關系

Question

我的MySQL數據庫包含兩個表： user和coupon （一對一的關系）。

我想選擇所有沒有優惠券的用戶並創建一個新的（隨機且唯一的）。

user TABLE:
___________________________________
|  id   |   name   |   coupon_id  |
-----------------------------------
   1        John         5
   2        Mary         (null)  // Need to create one coupon.
   3        Doe          2
   4        Max          (null)  // Need to create one coupon.
   5        Rex          1
   7        Bill         (null)  // Need to create one coupon.


coupon TABLE:
______________________________________________
|  id   |   code (random 6-chars - unique)   |
----------------------------------------------
   1        80k2ni
   2        0akdne
   5        nk03jd

---

快捷鍵：

選擇沒有優惠券的所有用戶： SELECT * from user WHERE coupon_id IS NULL;

生成一個隨機的6個字符串（MySQL）： LEFT(sha1(rand()), 6) 。

Answer 1

假設你不介意從下一個coupon_id繼續向上6，這可以按如下方式完成（參見SQL Fiddle Demo ）：

-- Declare and set variables
SET @id_for_insert = (SELECT MAX(`id`) FROM `coupon`);
SET @id_for_update = @id_for_insert;

-- Insert new coupons
INSERT INTO `coupon` (id, code)
SELECT @id_for_insert := @id_for_insert + 1, LEFT(SHA1(RAND()), 6)
FROM `user`
WHERE coupon_id IS NULL;

-- Update users that don't already have a coupon with the newly created coupons
UPDATE `user`
SET coupon_id = @id_for_update := @id_for_update + 1
WHERE coupon_id IS NULL;

Answer 2

這樣的事情可能是：

Insert into coupon 
select distinct id,LEFT(sha1(rand()), 6) 
from user WHERE coupon_id IS NULL

在此之后，您可以使用簡單的連接更新來更新用戶表中的優惠券。

Answer 3

//check if works inform me i will be glad
CREATE PROCEDURE `syncCouponUser` ()
BEGIN
  INSERT INTO coupon(id,code)//create coupons var no-used user ids
    select
    id,
    LEFT(sha1(rand())
    from user
    where coupon_id IS NULL ;

     UPDATE user //add coupons from coupon table with matches
    SET coupon_id=(
    select c.coupon_id
    from coupon c join user u 
    on c.id=u.id);

END